Convergence of $\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n^{1+1/n}}$

This series does not converge absolutely, by the limit comparison test

$$ \lim_{n\to\infty} \frac{1/n^{1+1/n}}{1/n} = \lim_{n\to\infty} \frac{1}{n^{1/n}} = 1 $$

with the divergent harmonic series $\sum \frac{1}{n}$.

But since

$$ \frac{1}{n^{1+1/n}} = \frac{1}{n} \exp\left( - \frac{\log n}{n} \right) = \frac{1}{n} + O\left(\frac{\log n}{n^2} \right), $$

the series converges conditionally by noting that

$$ \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{1+1/n}} = \sum_{n=1}^{\infty} (-1)^{n} \left( \frac{1}{n} + O\left(\frac{\log n}{n^2} \right) \right), $$

which is a sum of two convergent series. This is my favorite style of argument for proving that an alternating series with complicated term converges.

If you feel uncomfortable with the Big-Oh notation, then consider the function

$$ f(x) = x^{-1-\frac{1}{x}} = \exp\left( - \frac{x+1}{x}\log x \right). $$

Then by the logarithmic differentiation,

$$\frac{f'(x)}{f(x)} = -\frac{x+1-\log x}{x^2} < 0 $$

for large $x$ and thus $f(x)$ is a non-negative decreasing function. Since it is immediate that $f(x) \to 0$ s $x \to \infty$, the conclusion follows from the alternating series test that

$$ \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{1+1/n}} = \sum_{n=1}^{\infty} (-1)^{n} f(n) $$

converges.